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Assessment schedule
- Quiz 2 7:00 PM Fri 12 April 2019 (Week 7)
- Assignment 2 11:59 PM Sun 14 April 2019
- Quiz 3 7:00PM Fri 24 May 2019 (Week 12)
- Assignment 3 11:00PM Fri 24 May 2019
- Teaching Assistant
- Darcy Townsend [email protected]
1
Lecture 7:
Dynamic Programming I
Fast Fourier Transform
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Algorithmic Paradigms
- Greed. Build up a solution incrementally, myopically optimizing
some local criterion.
- Divide-and-conquer. Break up a problem into two sub-
problems, solve each sub-problem independently, and combine
solution to sub-problems to form solution to original problem.
- Dynamic programming. Break up a problem into a series of
overlapping sub-problems, and build up solutions to larger and
larger sub-problems.
4
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Dynamic Programming Applications
- Areas.
- Bioinformatics.
- Control theory.
- Information theory.
- Operations research.
- Computer science: theory, graphics, AI, systems, ….
- Some famous dynamic programming algorithms.
- Viterbi for hidden Markov models.
- Unix diff for comparing two files.
- Smith-Waterman for sequence alignment.
- Bellman-Ford for shortest path routing in networks.
- Cocke-Kasami-Younger for parsing context free grammars.
5
Fibonacci: Recursive Call Tree
Fibonacci: Recursive Call Tree
6
5 4
4 3 3 2
(^3 2 2 1) (^2) 1 1 0
2 (^1) 1 1 (^0) 1 0
(^1 )
let's leverage all the repeats!
0
Memoisation: Don't re-do unnecessary work!
- Memoization : Store previous results so that in future executions, you don’t have to recalculate them aka remember what you have already done!
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6
5 4
4 3 3 2
(^3 2 2 1) (^2) 1 1 0
(^2 1) 1 1 (^0) 1 0 (^1) 0
let's leverage all the repeats!
0
Memoisation: Don't re-do unnecessary work!
fib(n) if n is 1 or 2, return 1; if M[n] is not zero, return M[n]; M[n] = fib(n-1) + fib(n-2); return M[n];
Recall Interval Scheduling (Lecture 5)
- Interval scheduling.
- Input: Set of n jobs. Each job i starts at time s (^) j and finishes at time f (^) j.
- Two jobs are compatible if they don't overlap in time.
- Goal: find maximum subset of mutually compatible jobs.
Time 0 1 2 3 4 5 6 7 8 9 10 11
f g h
e
a b c d
...
- Theorem: Greedy algorithm [Earliest finish time] is optimal.
- Proof: (by contradiction)
- Assume greedy is not optimal, and let's see what happens.
- Let i 1 , i 2 , ... i (^) k denote the set of jobs selected by greedy.
- Let J 1 , J 2 , ... Jm denote the set of jobs in an optimal solution with i 1 = J 1 , i 2 = J 2 , ..., i (^) r = Jr for the largest possible value of r.
Recall Interval Scheduling (Lecture 5)
J 1 J 2 Jr
i 1 i 1 ir ir+
...
solution still feasible and optimal, but contradicts maximality of r.
Jr+
job ir+1 finishes before Jr+
Greedy:
OPT:
Exchange argument!
Weighted Interval Scheduling
- Weighted interval scheduling problem.
- Job j starts at s (^) j, finishes at f (^) j, and has weight or value vj.
- Two jobs compatible if they don't overlap.
- Goal: find maximum weight subset of mutually compatible jobs.
Time 0 1 2 3 4 5 6 7 8 9 10 11
f g h
e
a b c d
Unweighted Interval Scheduling Review
- Recall. Greedy algorithm works if all weights are 1.
- Consider jobs in ascending order of finish time.
- Add job to subset if it is compatible with previously chosen jobs.
- Observation. Greedy algorithm can fail if arbitrary weights
are allowed.
Time 0 1 2 3 4 5 6 7 8 9 10 11
b
a
weight = 999
weight = 1
doesn’t work
Dynamic Programming: Binary Choice
Notation. OPT(j) = value of optimal solution to the problem
consisting of job requests 1, 2, ..., j.
- Case 1: OPT selects job j.
- can't use incompatible jobs { p(j) + 1, p(j) + 2, ..., j - 1 }
- must include optimal solution to problem consisting of remaining compatible jobs 1, 2, ..., p(j)
- Case 2: OPT does not select job j.
- must include optimal solution to problem consisting of remaining compatible jobs 1, 2, ..., j-
OPT ( j ) =
0 if j = 0 max (^) { v (^) j + OPT ( p ( j )), OPT ( j −1)} otherwise
Example
maximum
